• Sep 3rd 2007, 08:17 PM
jam333
the derivate of y + y = 2 / (1+4e^2x)
y(-ln2)=pi/2
y=e^-x*tan^-1(2e^x)
i need to find out the intial value plz someone help me work it out
greatly appreciated

:(:(:(
• Sep 3rd 2007, 08:37 PM
Jhevon
Quote:

Originally Posted by jam333
the derivate of y + y = 2 / (1+4e^2x)
y(-ln2)=pi/2
y=e^-x*tan^-1(2e^x)
i need to find out the intial value plz someone help me work it out
greatly appreciated

:(:(:(

so nice of you to find the answer for us.

anyway, here is how to continue (you forgot the arbitrary constant by the way)

we have:

$\displaystyle y = e^{-x} \arctan (2 e^x) + Ce^{-x}$

we are told that:

$\displaystyle y(- \ln 2) = \frac {\pi}{2}$

this means:

$\displaystyle y(- \ln 2) = e^{ \ln 2} \arctan \left( \frac {2}{e^{\ln 2}} \right) + Ce^{\ln 2} = \frac {\pi}{2}$

$\displaystyle \Rightarrow 2 \arctan(1) + 2C = \frac {\pi}{2}$

$\displaystyle \Rightarrow 2 \cdot \frac {\pi}{4} + 2C = \frac {\pi}{2}$

$\displaystyle \Rightarrow C = 0$

$\displaystyle \Rightarrow y = e^{-x} \arctan (2 e^x)$
• Sep 3rd 2007, 09:25 PM
jam333
thanks helped me outalot:D